Why is $x^3-x^2+x$ injective?

Question

The function $$ f(x) = x^3-x^2+x$$ is injective (as seen on the graph), but by doing f(a) = f(b) I can't get to the point where a = b.

Can this be indicated in an analytical way (without graph) ?

Answer 1

I think the easiest way to see that this is injective is via calculus: if for some $a\not=b$ we had $f(a)=f(b)$, there would have to be some $c$ with $f'(c)=0$. But $f'(x)=3x^2-2x+1$, which has no (real) solutions.

Answer 2

It's monotonically increasing because $f'(x)=3x^2-2x+1=3(x-\frac{1}{3})^2+\frac{2}{3}>0$ so it has to be injective.

Answer 3

Without calculus, you can inspect the expression $f(a)-f(b)=0$ using $$ \begin{align} f(a)-f(b)&=(a^3-a^2+a)-(b^3-b^2+b)\\ &=(a-b)\left((a^2+ab+b^2)-(a+b)+1 \right)\\ &=\textstyle(a-b)\left[ \left(a+\frac{b-1}2\right)^2+\frac34\left( b-\frac13\right)^2+\frac23\right], \end{align} $$ and note that the second term on the right can never be zero.

Answer 4

Well, you should not give up hope on that $f(a)=f(b)$ approach :-):

$$f(a)=f(b)\\a^3-a^2+a=b^3-b^2+b\\a^3-b^3-(a^2-b^2)+(a-b)=0\\(a-b)(a^2+ab+b^2-a-b+1)=0$$ Which means that either $a=b$ or $a^2+ab+b^2-a-b+1=0$ Considering this as a quadratic with respect to a, we calculate it's discriminant as $$\Delta=(b-1)^2-4(b^2-b+1)=\\-3b^2+2b-3$$Which can be showed to be $<0$.

Thus we must have $a=b$.

Answer 5

We do it the "hard" way, using algebra only.

Let us calculate $f(a)-f(b)$. This is $$(a^3-b^3)-(a^2-b^2)+(a-b).$$ Factor out $a-b$ out of each term. We get $$(a-b)(a^2+ab+b^2-a-b+1).$$ We want to show that $a^2+ab+b^2-a-b+1$ is never $0$. It is easy to verify that it cannot be $0$ if $a=0$. So assume that $a\ne 0$. Then we can think of our expression as a quadratic in $a$.

The disciminant is $(b-1)^2-4(b^2-b+1)$, which simplifies to $$-3b^2+2b-3.$$ It is easy to verify that this is never $0$, and therefore always negative.

Answer 6

Let $f(a) = f(b)$, then $f(a)-f(b) = 0 \iff a^3-a^2+a -b^3+b^2-b = 0 \iff a-b = a^2(1-a) -b^2(1+b)$

But this means $2(a-b) = a^2(1-a) -b^2(1+b) + a - b = (f(a)-f(b)) = 0$, from which $a = b$ follows.

Answer 7

Suppose $f(a)=f(b)$ for some $a,b \in \mathbb{R}$. Then, since $f(x)$ is continuous and differentiable, by Rolle's Theorem there must be some point $c \in (a,b)$ with $f'(c)=0$. But one can check directly that the equation $f'(x)=0$ has no real solutions.

Answer 8

You can do it without derivatives. For $a>0$ we have$$\begin{align}f(x+a)-f(x)&=(x+a)^3-(x+a)^2+(x+a)-(x^3-x^2+x)\\&=3ax^2+3a^2x+a^3-2ax-a^2+a\\&=a(3(x+\tfrac a2-\tfrac13)^2+\tfrac{a^2}4+\tfrac23)\\&>0\end{align}$$